Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $a = \dfrac{5n^2 + 45n}{n^2 - 2n - 24} \times \dfrac{-5n + 30}{2n + 18} $
First factor the quadratic. $a = \dfrac{5n^2 + 45n}{(n - 6)(n + 4)} \times \dfrac{-5n + 30}{2n + 18} $ Then factor out any other terms. $a = \dfrac{5n(n + 9)}{(n - 6)(n + 4)} \times \dfrac{-5(n - 6)}{2(n + 9)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ 5n(n + 9) \times -5(n - 6) } { (n - 6)(n + 4) \times 2(n + 9) } $ $a = \dfrac{ -25n(n + 9)(n - 6)}{ 2(n - 6)(n + 4)(n + 9)} $ Notice that $(n + 9)$ and $(n - 6)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ -25n(n + 9)\cancel{(n - 6)}}{ 2\cancel{(n - 6)}(n + 4)(n + 9)} $ We are dividing by $n - 6$ , so $n - 6 \neq 0$ Therefore, $n \neq 6$ $a = \dfrac{ -25n\cancel{(n + 9)}\cancel{(n - 6)}}{ 2\cancel{(n - 6)}(n + 4)\cancel{(n + 9)}} $ We are dividing by $n + 9$ , so $n + 9 \neq 0$ Therefore, $n \neq -9$ $a = \dfrac{-25n}{2(n + 4)} ; \space n \neq 6 ; \space n \neq -9 $